Question: $f(t) = 6t^{2}-3(g(t))$ $h(n) = n-2(g(n))$ $g(n) = n$ $ h(f(-1)) = {?} $
First, let's solve for the value of the inner function, $f(-1)$ . Then we'll know what to plug into the outer function. $f(-1) = 6(-1)^{2}-3(g(-1))$ To solve for the value of $f$ , we need to solve for the value of $g(-1)$ $g(-1) = -1$ $g(-1) = -1$ That means $f(-1) = 6(-1)^{2}+(-3)(-1)$ $f(-1) = 9$ Now we know that $f(-1) = 9$ . Let's solve for $h(f(-1))$ , which is $h(9)$ $h(9) = 9-2(g(9))$ To solve for the value of $h$ , we need to solve for the value of $g(9)$ $g(9) = 9$ $g(9) = 9$ That means $h(9) = 9+(-2)(9)$ $h(9) = -9$